CSE322: Finite Automata from Regular Expressions

CSE322: Construction of Finite Automata Equivalent to Regular Expressions

• The subset method constructs a finite automaton equivalent to a regular expression.
• This method involves two steps.
• Step 1: Construct a transition graph (transition system) equivalent to the given regular expression using A-moves. This is done using Theorem 5.2.
• Step 2: Construct the transition table for the transition graph.

Problem Example

• Construct the finite automaton equivalent to the regular expression (0 + 1)(00 + 11)(0 + 1).

Solution Steps (illustrated with diagrams)

• Diagrams (a), (b), (c), and (d) show the construction process, progressing from an initial expression to a more complex automaton. Each step refines the structure to effectively recognize the given regular expression.

Equivalence of Two Finite Automata

• Two finite automata are equivalent if they accept the same set of strings over the input Σ.
• If two finite automata are not equivalent, there exists a string w over Σ such that one automaton reaches a final state while the other reaches a non-final state when presented with w.

Cases for Determining Equivalence

• Case 1: If a pair (q, q') is encountered where 'q' is a final state in one automaton and 'q'' is a non-final state in the other (or vice versa), then the automata are not equivalent. Construction is terminated.
• Case 2: If in the construction, no new elements appear in subsequent columns that are not already present in the first column, then the automata are equivalent. Construction is terminated.

Example Problem and Solution

• Consider two DFAs (M and M') over {0, 1} (as depicted in Figure 5.23). The goal is to determine if they are equivalent.
• The solution demonstrates the process of comparing the states of the two finite automata to determine if they are equivalent step-by-step.
• The final conclusion is that the automata are equivalent.

Equivalence of Two Regular Expressions

• Two regular expressions P and Q are equivalent if they generate the same set of strings.
• The equivalence of regular expressions is equivalent to the equivalence of the corresponding finite automata.

Proof Example

• Prove (a + b)* = a*(ba*)*.
• The provided solution outlines how to prove this equivalence by showing that the transition systems generated by both expressions are equivalent/similar. Diagrams (Figures 5.25 and 5.26) illustrate the relevant transition systems.

Closure Properties of Regular Sets

• Regular sets exhibit closure under various operations:
  • Set union
  • Concatenation
  • Closure (iteration)
  • Transpose
  • Set intersection
  • Complementation

This quiz focuses on the construction of finite automata that are equivalent to given regular expressions. It outlines the subset method, detailing the two steps necessary to create transition graphs and tables from regular expressions. Additionally, it covers the equivalence of finite automata, emphasizing the importance of recognizing the same set of strings.